Page 271 - 48Fundamentals of Compressible Fluid Mechanics

13.4. SOLUTION OF MACH ANGLE 233

Example 13.5:
M y w
A supersonic flow approaching a very long two di- A ∗
10 ◦
mensional bland wedge body and creates a detached M y s
shock at Mach 3.5 (see Figure 13.19). The half wedge
angle is . What is the requited “throat” area ratio
to achieve acceleration from subsonic region to super- Fig. 13.19: Schematic for ex-
sonic region assuming one-dimensional flow. ample 13.5

SOLUTION
The detach shock is a normal shock and the results are

5

4
4
5

7
6
6
7
9 9 9
9
9
+ "
Now utilizing the isentropic relationship for yields 78 78 "-9

4
5
7
6
7
7 8
5
"-+ -"
" 9 +
-+1"1+-+
90
9-)
9 9
Thus the area ratio has to be 1.4458. Note that the pressure after the weak shock

is irrelevant to area ratio between the normal shock to the “throat” according to the

standard nozzle analysis.

6 8
: 7 8
8
Example 13.6:
D 4 Slip Plane
The effects of double
wedge were explained in B P 3 = P 4
government web site as weak 3
oblique weak
shown in figure 13.20. shock oblique
shock E
Adopted this description or expension
wave
and assumed that turn of M 1 1 2
is made out of two equal 0 C
(see Figure A
+
13.20). Assume that there
are no boundary layers and Fig. 13.20: Schematic of two angles turn with two weak
angles of
all the shocks are weak shocks
and straight. Carry the
.
Find the required angle of

calculation for
shock BE. Then, explain why this description has internal conflict.

SOLUTION
The shock BD is an oblique shock which response to total turn of + . The condition
for this shock are: